Rewrite the equation by completing the square. $4 x^{2} -4 x +1 = 0$ $(x + $
Solution: $\begin{aligned} 4 x^2 -4 x +1&=0 \\\\ 4 x^2 -4 x &=-1 \\\\ x^2 - x&=-\dfrac{1}{4} \end{aligned}$ Now we want to complete $x^2 - x$ into a perfect square. To do that, we should add $\left(\dfrac{{-1}}{2}\right)^2={\dfrac{1}{4}}$ to it: $x^2{-1}x + {\dfrac{1}{4}}=\left(x -\dfrac{1}{2} \right)^2$ $\begin{aligned} x^2 - x&=-\dfrac{1}{4} \\\\ x^2 - x + {\dfrac{1}{4}}&=-\dfrac{1}{4} + {\dfrac{1}{4}} \\\\ \left(x -\dfrac{1}{2} \right)^2&=0 \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x -\dfrac{1}{2} \right)^2=0$